Question

What is the value of $\left[{\left({\left(x^{y+1}\right)}^{\frac{y^2}{y^2-1}}\right)}^{1-\frac{1}{y}}\right]$?

• A. $xy$
• B. $x^y$
• C. $y^x$
• D. $\frac{y}{x}$

Answer 1

The correct option is B $x^y$

Consider $\left[{\left({\left(x^{y+1}\right)}^{\frac{y^2}{y^2-1}}\right)}^{1-\frac{1}{y}}\right]$ and simplify it as follows:

$\left[{\left({\left(x^{y+1}\right)}^{\frac{y^2}{y^2-1}}\right)}^{1-\frac{1}{y}}\right]=\left[{\left({\left(x^{y+1}\right)}^{\frac{y^2}{(y+1)(y-1)}}\right)}^{1-\frac{1}{y}}\right]=\left[{\left(x^{\frac{y^2}{y-1}}\right)}^{\frac{y-1}{y}}\right]=x^{\left(\frac{y^2}{y-1}\right)\left(\frac{y-1}{y}\right)}=x^y$

Hence, $\left[{\left({\left(x^{y+1}\right)}^{\frac{y^2}{y^2-1}}\right)}^{1-\frac{1}{y}}\right]=x^y$